Question

Calculate the force between two Protons, if they are 4.0 × 10-15 meters away from
each other. (Where, charge of Proton is 1.6 x 10-19 Coulomb.)​

Answer 1

Answer:

14.4 N

Explanation:

Force for static electricity is -

• $\dfrac{1}{4\pi\epsilon} \dfrac{q_1.q_2}{r^2}$

Here,

r = distance between them = $4.0 \times 10^{-15}$

$q_1 = q_2$ = charge on particles, here, both are same. So

$q_1 = q_2 = 1.6\times10^{-19}$

We know, $\dfrac{1}{4\pi\epsilon}=9 \times 10^9$

Hence,

$Force=9 \times 10^9 \times \frac{1.6\times10^{-19}\times1.6\times10^{-19}}{(4.0\times10^{-15})^2}\\\\\\Force=9 \times 10^9 \times \frac{1.6\times10^{-19}\times1.6\times10^{-19}}{(16\times10^{-30}}\\\\\\ Force = 9\times \dfrac{1.6\times1.6}{16}\times10^{-19-19+30+9}\\\\\\ Force = 1.44 \times 10^{1}$

Force = 14.4 N